So, I was sitting around one day fiddling with an old Groove Driver, and a question popped into my head...which driver is the most efficient? The Groove, Rckstr's driver, or the FlexDrive V5? The V5 is handy, since it's a boost circuit, but that means it also has to modify the electricity, where the other two only have to regulate it. You (at least in theory) could use all three with the same source voltage (5.5), set each one to the same current, and then used something like an oscilloscope or even just a multi-meter/ampremeter, in series to measure the current being sucked off the battery, then you might be able to figure out which one is most efficient by seeing which one uses the least amount of current.

Are there any huge flaws in my idea, or does it appear to be a sound theory?

If anyone can shed any light on this and theorize (with an explanation) for which one is the most efficient, I'd like to know. My first guess is the Groove driver, since it has a lower drop-out voltage than the Rckstr one, but the Rckstr driver has fewer components which could mean more efficiency...you're all free to give your two cents, but please explain why you think whatever you think about the drivers.

I'd say any linear driver like the rkcstr or DDL will be more efficient than a buck/boost driver like the flexdrive. The rkcstr and DDL will have nearly identical efficiency, since the rkcstr is pretty much just a micro version of the DDL. The groove driver is also linear, and will likely be as efficient as the rkcstr/DDL drivers.

The flexdrive draws extra power to boost voltage as part of it's function, so it will not be as efficient, but it does tend to be a lot more handy for certain builds.

I'd say any linear driver like the rkcstr or DDL will be more efficient than a buck/boost driver like the flexdrive. The rkcstr and DDL will have nearly identical efficiency, since the rkcstr is pretty much just a micro version of the DDL. The groove driver is also linear, and will likely be as efficient as the rkcstr/DDL drivers.

The flexdrive draws extra power to boost voltage as part of it's function, so it will not be as efficient, but it does tend to be a lot more handy for certain builds.

Incorrect.

The linear drivers are NOT efficient. They take the extra voltage and turn it into heat.
The DC boost diodes however are very efficient. Atleast 80%. Sometimes over 90%.

So basically, the best efficiency a linear driver gets for a bluray is 69%.
Then the efficiency drops as voltage is increased since the extra power is converted into heat. This is why linear drivers get hot.

The DC boost drivers however DO NOT CONVERT EXTRA POWER TO HEAT.
This is because they are given lower voltage than output voltage.

So say a flexdrive has an input of 3V and needs to output 5V @ 100ma to drive a PHR.

So basically, the best efficiency a linear driver gets for a bluray is 69%.
Then the efficiency drops as voltage is increased since the extra power is converted into heat. This is why linear drivers get hot.

The DC boost drivers however DO NOT CONVERT EXTRA POWER TO HEAT.
This is because they are given lower voltage than output voltage.

So say a flexdrive has an input of 3V and needs to output 5V @ 100ma to drive a PHR.

However since they are about 90% efficient. The Input current will actually be about 185ma.
So in this case at a efficiency of 90%, the boost driver takes in 3v@185ma and outputs 5V @ 100ma

Hope this made sence

Feel free to point out any errors.

That does make a lot of sense. My concern is simply whether or not a boost circuit that small will be able to retain that much efficiency. But your logic seems certainly sound. What I'm curious about is if you could have a (virtually) zero drop out voltage linear circuit, or is the only way to get an effective zero dropout by using a boost circuit that can modify the electricity coming in?

One of the reasons I developed the FlexDrive was to overcome the poor efficiency and wasted heat of the linear drivers. And the fact that it can do this at near 90% efficiency and be so small is a big reason why it's so cool.
Even if you had a 0V dropout linear regulator, it would STILL be much less efficient than a FlexDrive over battery life because the input battery voltage isn't constant. As your battery dropped from 4.2V to 3V while driving a red, you still would be wasting (VBatt-VDiode)*current as heat into your laser.

the flexdrive is a buck/boost driver. Buck meaning that it converts the excess voltage to current instead of heat. Which is different from just calling it a boost driver, because a boost only driver will still boost the voltage and still convert excess voltage to heat.

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